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PODRĘCZNIK PHP 5.x, 4.x, 3.x - częściowo spolszczony / źródło: www.php.net
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Zwracanie przez-referencję jest użyteczne gdy chcesz użyć funkcji do
znalezienia zmiennej do której referencją powinna być dowiązana. Zwracając
referencje używaj tej składni:
W tym przykładzie, będzie zmieniona własność obiektu zwróconego przez
funkcję znajdz_zmienna, a nie własność kopii obiektu,
jak to by było bez użycia składni referencji.
Notatka:
Inaczej niż przy przekazywaniu parametrów, tutaj musisz używać
& w obu miejscach - by wskazać, że zwracasz
przez-referencję, a nie kopię jak normalnie, i by zwrócić uwagę, że
dla zmiennej $foo powinno być użyte powiązanie przez
referencję, a nie zwykłe przypisanie.
User Contributed Noteswarhog at warhog dot net
13-Dec-2005 09:04
firstly a note on the post below: technically correct that -> "to get a reference to an exsisting class and it's properties" should be "...to an existing object..."
in PHP5 it's senseless to return objects by reference.. let's say you have, as in the post below, a class which should return a reference to its own instance (maybe it's been created using the singleton pattern..), than it's no problem to simply return that variable holding the instance.
In PHP5 variables holding objects are always references to a specific object, so when you return (=copy) a variable "having" an object you in fact return a reference to that object.
Because of that behaviour, which is very common for many programming languages, especially those, which are to be compiled (due to memory problems and so on), you have to use the clone-function which was introduced in PHP5.
Here an example to underline what i mean:
<?php
class sample_singleton
{
protected static $instance = NULL;
private function __construct()
{ }
public static function create()
{ self::$instance = new sample_singleton(); }
public static function getInstance()
{ return self::$instance; }
public function yea()
{ echo "wuow"; }
}
sample_singleton::create();
$singleton = sample_singleton::getInstance();
$singleton->yea();
?>
Note some more (maybe) strange behaviour: although $instance is private you can have a reference pointing on it. Maybe that does not seem strange to you in any way, but maybe you wondered as i did : )
The code posted here was just a sample to illustrate my posting, for using the singleton pattern please look it up in the manual. I just used this cause it was the simplest example which went in my mind right know.
willem at designhulp dot nl
09-Oct-2005 01:54
There is an important difference between php5 and php4 with references.
Lets say you have a class with a method called 'get_instance' to get a reference to an exsisting class and it's properties.
<?php
class mysql {
function get_instance(){
if(empty($_ENV['instances']['mysql'])){
$_ENV['instances']['mysql'] = new mysql;
}
$ref = &$_ENV['instances']['mysql'];
return $ref;
}
}
?>
Now to get the exsisting object you can use
mysql::get_instance();
Though this works in php4 and in php5, but in php4 all data will be lost as if it is a new object while in php5 all properties in the object remain.
obscvresovl at NOSPAM dot hotmail dot com
24-Dec-2004 10:09
An example of returning references:
<?
$var = 1;
$num = NULL;
function &blah()
{
$var =& $GLOBALS["var"]; $var++;
return $var;
}
$num = &blah();
echo $num; blah();
echo $num; ?>
Note: if you take the & off from the function, the second echo will be 2, because without & the var $num contains its returning value and not its returning reference.
hawcue at yahoo dot com
17-Mar-2004 04:58
Be careful when using tinary operation condition?value1:value2
See the following code:
$a=1;
function &foo()
{
global $a;
return isset($a)?$a:null;
}
$b=&foo();
echo $b; // shows 1
$b=2;
echo $a; // shows 1 (not 2! because $b got a copy of $a)
To let $b be a reference to $a, use "if..then.." in the function.
contact at infopol dot fr
12-Feb-2004 06:36
A note about returning references embedded in non-reference arrays :
<?
$foo;
function bar () {
global $foo;
$return = array();
$return[] =& $foo;
return $return;
}
$foo = 1;
$foobar = bar();
$foobar[0] = 2;
echo $foo;
?>
results in "2" because the reference is copied (pretty neat).
zayfod at yahoo dot com
03-Dec-2003 06:23
There is a small exception to the note on this page of the documentation. You do not have to use & to indicate that reference binding should be done when you assign to a value passed by reference the result of a function which returns by reference.
Consider the following two exaples:
<?php
function & func_b ()
{
$some_var = 2;
return $some_var;
}
function func_a (& $param)
{
$param = & func_b();
}
$var = 1;
func_a($var);
?>
The second example works as intended:
<?php
function & func_b ()
{
$some_var = 2;
return $some_var;
}
function func_a (& $param)
{
$param = func_b();
}
$var = 1;
func_a($var);
?>
(Experienced with PHP 4.3.0)
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